Swap Stability in Schelling Games on Graphs

09/05/2019 ∙ by Aishwarya Agarwal, et al. ∙ 0

We study a recently introduced class of strategic games that is motivated by and generalizes Schelling's well-known residential segregation model. These games are played on undirected graphs, with the set of agents partitioned into multiple types; each agent either occupies a node of the graph and never moves away or aims to maximize the fraction of her neighbors who are of her own type. We consider a variant of this model that we call swap Schelling games, where the number of agents is equal to the number of nodes of the graph, and agents may swap positions with other agents to increase their utility. We study the existence, computational complexity and quality of equilibrium assignments in these games, both from a social welfare perspective and from a diversity perspective.

READ FULL TEXT VIEW PDF
POST COMMENT

Comments

There are no comments yet.

Authors

page 1

page 2

page 3

page 4

This week in AI

Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.

1 Introduction

Segregation is observed in many communities; people tend to group together on the basis of politics, religion, or socioeconomic status. This phenomenon has been extensively documented in residential metropolitan areas, where people may select where to live based on the racial composition of the neighborhoods. To formalize and study how the motives of individuals lead to residential segregation, Schelling (1969, 1971) proposed the following simple, yet elegant model. There are two types of agents who are to be placed on a line or a grid. An agent is happy with her location if at least a fraction of the agents within a certain radius are of the same type as her. Happy agents do not want to move, but unhappy agents are willing to do so in hopes of improving their current situation. Schelling described a dynamic process where at each step unhappy agents jump to random open positions or swap positions with other randomly selected agents, and showed that it can lead to a completely segregated placement, even if the agents themselves are tolerant of mixed neighborhoods ().

Over the years, Schelling’s work became very popular among researchers in Sociology and Economics, who proposed and studied numerous variants of his model, mainly via agent-based simulations; see the paper of Clark and Fossett (2008) and references therein for examples of this approach. Variants of the model have also been rigorously analyzed in a series of papers (Young, 2001; Zhang, 2004a; Brandt et al., 2012; Barmpalias et al., 2014; Bhakta et al., 2014; Barmpalias et al., 2015; Immorlica et al., 2017)

, which showed that the random behavior of the agents leads with high probability to the formation of large monochromatic regions.

While all these papers focused on settings where the agents’ behavior is random, it is more realistic to assume instead that the agents are strategic and move only when they have an opportunity to improve their situation. So far, only a few papers have followed such a game-theoretic approach. In particular, Zhang (2004b) considered a game where the agents optimize a single-peaked utility function, and very recently, Chauhan et al. (2018), Elkind et al. (2019) and Echzell et al. (2019) studied strategic settings that are closer to the original model of Schelling, but capture more that two agent types and richer graph topologies.

In particular, Chauhan et al. (2018) study a setting with two types of agents, who have preferred locations, and can either swap with other agents or jump to empty positions. For a given tolerance threshold , each agent’s primary goal is to maximize the fraction of her neighbors that are of her own type as long as this fraction is below (with all fractions above being equally good); her secondary goal is to be as close as possible to her preferred location. For both types of games (swap and jump), Chauhan et al. identify values of for which the best response dynamics of the agents leads to an equilibrium when the topology is a ring or a regular graph. Echzell et al. (2019) strengthen these results and extend them to more than two agent types, as well as study the complexity of computing assignments that maximize the number of happy agents.

Elkind et al. (2019) consider a similar model with types; however, they treat agents’ location preferences differently from Chauhan et al. Namely, in their model each agent is either stubborn (i.e., has a preferred location and is unwilling to move) or strategic (i.e., aims to maximize the fraction of her neighbors that are of her own type; this corresponds to setting in the model of Chauhan et al.). They focus on jump games, i.e., games where agents may jump to empty positions, and analyze the existence and complexity of computing Nash equilibria, as well as prove bounds on the price of anarchy Koutsoupias and Papadimitriou (1999) and the price of stability Anshelevich et al. (2008).

1.1 Our Contribution

We combine the two approaches by considering swap games in the model of Elkind et al. (2019). That is, we assume that the number of agents is equal to the number of nodes in the topology, and two agents can swap locations if each of them prefers the other agent’s location to her own. We begin by studying the existence of equilibrium assignments. While such assignments exist for highly structured topologies, we show that they may fail to exist in general, even for simple topologies such as trees. Moreover, we show that deciding whether an equilibrium exists is NP-complete. We also study the quality of assignments in terms of their social welfare: we prove bounds on the price of anarchy and the price of stability for many interesting cases, and show that computing an assignment with high social welfare is NP-complete; the latter result complements the result of Elkind et al. in that it applies to the case where the number of agents equals the number of nodes in the topology.

Interestingly, even though the goal of Schelling’s work was to study integration, neither his paper nor the follow-up work have defined a quantitative measure of diversity. To fill this gap, we initiate the formal study of diversity of equilibrium assignments, by defining a new benchmark called degree of integration. This benchmark counts the number of agents who are exposed to agents of other types, i.e., have at least one neighbor of a different type. We show that even the best equilibria are often much less diverse than the maximally diverse assignments, by bounding the price of anarchy and stability with respect to this measure. We also show that maximizing diversity is computationally hard.

1.2 Further Related Work

As mentioned above, Schelling’s model has been studied extensively both empirically and theoretically. For an introduction to the model and a survey of its many variants, we refer the reader to the book of Easley and Kleinberg (2010), and the papers by Brandt et al. (2012) and Immorlica et al. (2017). Besides the closely related papers by Chauhan et al. (2018), Elkind et al. (2019) and Echzell et al. (2019), another work that is similar in spirit is a recent paper by Massand and Simon (2019), who study swap stability in games where a set of items is to be allocated among agents who are connected via a social network, so that each agent gets one item, and her utility depends on the items she and her neighbors in the network get; however, their results are not directly applicable to our setting. Also, Schelling games share a number of properties with hedonic games (Drèze and Greenberg, 1980; Bogomolnaia and Jackson, 2002), and in particular, with fractional hedonic games (Aziz et al., 2019) and hedonic diversity games (Bredereck et al., 2019). However, a fundamental difference between hedonic games and Schelling games is that in the former agents form pairwise disjoint coalitions, while in the latter the neighborhoods of different nodes of the topology may overlap.

2 Preliminaries

A -swap game is given by a set of agents partitioned into pairwise disjoint types , and an undirected simple connected graph with , called the topology. We often identify types with colors: e.g., in a -swap game each agent is either red () or blue (

). The agents are also classified as either

strategic or stubborn. We denote by the set of strategic agents and by the set of stubborn agents, so that . Stubborn agents never move away from the nodes they occupy, while a strategic agent aims to maximize her personal utility, and is willing to swap positions with other agents to achieve this.

Given an agent , we refer to all other agents in as friends of and denote the set of ’s friends by . Each agent occupies some node of the topology so that for every pair of agents

. The vector

that lists the locations of all agents is called an assignment. Given an assignment , we denote by the agent that occupies node , that is, .

Given an assignment , let be the set of neighbors of agent . The utility of a stubborn agent is independent of the assignment; e.g., we can set for each . The utility of a strategic agent for assignment is

Observe that, since , every node is occupied by some agent, and therefore for every .

For every assignment , let be the assignment that is obtained from by swapping the positions of agents and : for every , and . Agents and swap positions if and only if they both strictly increase their utility: and . Clearly, agents of the same type cannot both increase their utilities by swapping, so swaps always involve agents of different types. An assignment is an equilibrium if no pair of agents want to swap positions. That is, is an equilibrium if and only if for every we have or . We denote the set of all equilibrium assignments of the -swap game by .

For every assignment , we define two benchmarks that aim to capture, respectively, the agents’ happiness and the societal diversity. The first one is the well-known social welfare, defined as the total utility of all strategic agents:

Our second benchmark is the degree of integration: we say that an agent is exposed is she has at least one neighbor of a different type, and count the number of exposed agents:

Note that we ignore the stubborn agents in the definitions of our benchmarks, as their utility is always the same and they never want to move somewhere else.

For , let be the optimal assignment in terms of the benchmark for a given game . The price of anarchy (PoA) in terms of the benchmark is the worst-case ratio (over all -swap games such that ) between the optimal performance (among all assignments) and the performance of the worst equilibrium assignment. Similarly, the price of stability (PoS) in terms of is the worst-case ratio between the optimal performance and the performance of the best equilibrium:

For readability, we refer to the quantity as the social price of anarchy and to as the integration price of anarchy, and use similar language for the price of stability.

3 Existence of Equilibria

We begin by discussing the existence of equilibria in swap games. The work of Echzell et al. (2019) implies that at least one equilibrium assignment exists when the topology is a regular graph. Furthermore, using a potential function similar to the one used by Elkind et al. (2019) for jump games, we can show that equilibria always exist when the topology is a graph of maximum degree ; we omit the details. Our first result is a proof of non-existence of equilibria for every for general topologies.

Figure 1: The topology of the -swap game considered in the proof of Theorem 3.1, and an assignment that corresponds to the last case in the analysis; it is not an equilibrium since the red agent at node and the blue agent at node would like to swap.
Theorem 3.1.

For every , there exists a -swap game that does not admit an equilibrium assignment, even when all agents are strategic and the topology is a tree.

Proof.

We start with the easiest case of . Consider a -swap game with strategic agents: red agents and blue agents. The topology is a tree with a root node , which has three children nodes (set ), each of which has two children of its own (set ); see Figure 1. Suppose for the sake of contradiction that this game admits an equilibrium assignment .

Since and there are only two types of agents, at least two nodes in , say and , must be occupied by agents of the same type, say red. Now assume that nodes (a child of ) and (a child of ) are occupied by blue agents. Then the red agent and the blue agent can swap positions to increase their utility from strictly smaller than and to and positive, respectively. Therefore, for at least one of these nodes (say, ) it must be the case that both of its children are occupied by red agents; as there are only five red agents, it follows that at least one of the children of , say , is occupied by a blue agent.

If node is occupied by a blue agent, then the red agent and the blue agent can both increase their utility by swapping. Hence, node must be occupied by a red agent (see Figure 1). However, this assignment is not an equilibrium either, since the red agent and the blue agent have an incentive to swap.

For , consider a -swap game with agents such that there are agents of type , for every . The topology is a tree whose nodes are distributed over three layers, just like in the case . Specifically, there is a root node , which has a set of children. Each node in has a set of children leaf nodes; let . Next, we will argue about the structure of assignments that cannot be equilibria.

Lemma 3.2.

An assignment according to which any two nodes are occupied by agents of the same type cannot be an equilibrium in case agents of some type , simultaneously occupy nodes of and .

Proof.

Let be an assignment according to which nodes are occupied by agents of type , and there exist nodes and , which are occupied by agents of some type , with . Clearly, agent has utility strictly less than , while agent has utility . Therefore, they would like to swap positions in order to increase their utility to and positive, respectively. ∎

Lemma 3.3.

An assignment according to which agents of every type occupy nodes of cannot be an equilibrium.

Proof.

Let be an assignment according to which at least one agent of every type occupies some node of . Without loss of generality, assume that the agent is of type . We now distinguish between the following two cases.

  • An agent of type located at node has a neighbor of type , located at some node . Then, by the assumption of the lemma, there exists at least one agent of type located at some node , and therefore agents and would like to swap positions in order to increase their utility from strictly less than and to and positive, respectively.

  • For every agent of type located at some node , all agents occupying the nodes of are of type . Since is occupied by an agent of type , there are other agents of type that can completely fill up at most subtrees of (since each of them consists of nodes). Consequently, there are at least agents of type located at leaf nodes in other subtrees of .

    Now, assume that one of these agents of type occupies a node such that is occupied by an agent of type , with . We will now argue that there must exist another agent of type located at some node . Assume otherwise that there is no such agent. Then, the remaining agents of type occupy leaf nodes of the tree. By Lemma 3.2, such agents located in different subtrees of cannot be connected to agents of the same type. Hence, to cover all these agents of type , agents of different types need to occupy the root nodes of the corresponding subtrees of . However, there are only types left (different than and ). Consequently, there must exist another agent of type that occupies some node . As a result, agents and have incentive to swap positions in order to increase their utility from and strictly less than to positive and , respectively.

This completes the proof of the lemma. ∎

Lemma 3.4.

An assignment according to which there exists a type such that no agent of this type appears at nodes of cannot be an equilibrium.

Proof.

Let be an assignment according to which no agent of type appears at the nodes of . We first deal with the case . Observe that there are at least agents of type that must occupy nodes of ; one agent of type may occupy . By Lemma 3.2, agents of the same type that are located in different subtrees of cannot be connected to agents of the same type. Hence, to cover all these agents of type , agents of different types must occupy the root nodes of the corresponding subtrees of , which is impossible.

For , if is not occupied by an agent of type , then all agents of this type must occupy nodes of , and the same argument as above leads to a contradiction. Hence, assume that is of type . Since and no agent of type appears at the nodes of , at least three nodes of must be occupied by agents of the same type. Let , and assume that nodes and are occupied by agents of type . If an agent of type occupies some node for any , then agent , has incentive to swap positions with agent to increase both of their utilities from strictly less than and to and positive. Hence, no agent of type can be located at the nodes of . Clearly, if agent or is of type , for the same reason, no agent of type can be located at the corresponding leaf nodes. Hence, both and must be occupied by agents of the third type and all leaf nodes must be occupied by the remaining agents of type . However, this clearly cannot be an equilibrium assignment, since agent would like to swap with any agent in . ∎

By Lemmas 3.3 and 3.4, we conclude that no assignment can be an equilibrium. ∎

The topology used in the proof of Theorem 3.1 for the case is utilized as a subgraph in the proof of the following theorem, to show that the problem of deciding whether an equilibrium exists is computationally hard.

Theorem 3.5.

For every , it is NP-complete to decide whether a given -swap game admits an equilibrium.

Proof.

Membership in NP is immediate: we can verify whether a given assignment is an equilibrium by simply checking if there exists a pair of agents that would like to swap positions. To prove NP-hardness, we give a reduction from the Clique problem, which in known to be NP-complete. An instance of Clique consists of an undirected connected graph and an integer ; it is a yes-instance if contains a complete subgraph of size . Without loss of generality, we assume that .

Given an instance of Clique with , we will construct a -swap game as follows (the reduction can be extended to any by adding stubborn agents of different types). Let denote the degree of node in , and set .

  • There are strategic red agents and strategic blue agents; all other agents are stubborn, and will be defined in conjunction with the topology.

  • The topology consists of three components , and . These are connected to each other via stubborn agents, and are defined as follows:

    • To define , let be a set of nodes for each . Then, and . For every , nodes of are occupied by stubborn red agents, while the remaining nodes are occupied by stubborn blue agents. Observe that every node of has degree .

    • The subgraph is a complete bipartite graph with and . Out of the nodes of , nodes are occupied by stubborn red agents, while the remaining nodes are occupied by stubborn blue agents.

      Hence, a strategic red agent occupying a node of has utility . Similarly, a strategic blue agent has utility .

    • To define , let be the graph used in the proof of Theorem 3.1, for which there is no equilibrium assignment; see Figure 1. For every node of degree , let be a set of nodes such that of these nodes are occupied by stubborn red agents, while the remaining nodes are occupied by stubborn blue agents. For every of degree , let be a set of nodes such that of these nodes are occupied by stubborn red agents, while the remaining nodes are occupied by stubborn blue agents. Then, and .

      One can easily verify that the utility of a strategic agent (red or blue) occupying a node of is at least and at most .

Now, assume that has a clique of size , and let be the assignment in which the strategic red agents occupy the nodes of the clique, and the strategic blue agents occupy the remaining nodes. Each strategic red agent is connected to other red agents (strategic and stubborn) in , and thus has utility

Clearly, since and , no strategic red agent would be willing to swap positions with another strategic agent in or . By swapping positions with a blue agent within , a strategic red agent would still have at most friends, and since every node in has the same degree, her utility cannot be improved. Hence, no strategic red agent has a profitable deviation, and is an equilibrium.

Conversely, assume that does not contain a clique of size , and for the sake of contradiction also assume that there is an equilibrium assignment .

Suppose that some strategic red agents are located in . It cannot be the case that each of them is adjacent to other strategic red agents, as this would mean that the nodes they occupy form a clique of size . Hence, at least one of these agents, say agent , is adjacent to at most strategic red agents. Since every node of has degree and every node is adjacent to stubborn red agents, the utility of is

We have that and , and hence agent has incentive to move to or . On the other hand, the utility that a strategic blue agent that is currently located in or can obtain by swapping positions with is

Since and , agent also has an incentive to swap positions with agent , and hence cannot be an equilibrium assignment. Therefore, no strategic red agent is located in .

Similarly, observe that and , meaning that strategic red agents would prefer to be in , while strategic blue agents would prefer to be in . Thus, for to be an equilibrium assignment, it must be the case that all if a node of is not occupied by a stubborn agent, it is occupied by a strategic red agent. As a result, there are strategic red and strategic blue agents in . However, similarly to the proof of Theorem 3.1, we can argue that there is no equilibrium assignment for these agents in ; we omit the details here. Since we have exhausted all possibilities, it follows that if does not have a clique of size , then there is no equilibrium assignment. ∎

4 Social Welfare

Here, we consider the efficiency of equilibrium assignments in terms of social welfare, and bound the social price of anarchy and stability for many interesting cases. We restrict our attention to games consisting of only strategic agents and such that there are at least two agents per type. Swap games with stubborn agents or strategic agents that are unique of their type can easily be seen to have unbounded social price of anarchy.111For any , consider a -swap game with a star topology and types of agents such that there is only one red strategic agent, while the other types consist of at least two strategic agents and of some stubborn ones located at peripheral nodes. The assignment according to which the red agent occupies the center node is an equilibrium with social welfare, while any assignment such that the center node is occupied by a non-red agent has positive social welfare.

We start with the social price of anarchy of -swap games, and consider the general case (given the above restrictions) and the case where each type consists of the same number of agents.

Figure 2: The topology and the equilibrium assignment that characterize the case where some agents have zero utility in the proof of part (i) in Theorem 4.1. The big red square represents any arbitrary connected component whose nodes are occupied only by red agents.
Theorem 4.1.

The social price of anarchy of -swap games with only strategic agents is

  • if there are at least two agents of each type, and

  • between and if each type consists of the same number of agents.

Proof.

We prove each part separately.

Part(i). For the lower bound, consider a -swap game with red and blue agents who are to be positioned on the nodes of a star topology. Then, the assignment according to which one of the blue agents occupies the central node is an equilibrium with social welfare , while the optimal assignment is such that the central node is occupied by a red agent for a social welfare of .

For the upper bound, consider any -swap game with agents such that there are red and blue agents, and let be any equilibrium assignment of this game. If there is any red agent with zero utility in , then it cannot be the case that this agent is connected to all blue agents. If this were the case, then since there is another red agent and the graph is connected, at least one blue agent must be connected to this red agent, get utility strictly less than , and have incentive to swap positions with agent so that they both increase their utility. Hence, in order for to be an equilibrium in the presence of getting zero utility, any blue agent not connected directly to must get utility so that she does not want to swap with . Consequently, . In the case where every agent has positive utility in , since the graph is connected, it must be the case that every agent get utility at least , and therefore again . Now the bound follows since the optimal social welfare is at most .

Part (ii). Consider any -swap game with agents such that there are red and blue agents, and let be any equilibrium assignment of this game.

First assume that some agents get zero utility. Clearly, it cannot be the case that agents of both types have zero utility, since otherwise such agents would have incentive to swap and increase their utility from zero to . So, assume that at least one blue agent has zero utility, while all red agents have positive utility. Then, it must be the case that all blue agents are connected to a single red agent , and all other red agents have utility . To see this, assume otherwise that there exists a red agent that is also connected to some blue agents, and thus has utility strictly smaller than . Then, agent would have incentive to swap with agent to increase the utility of both of them, contradicting the fact that is an equilibrium. Since agent has positive utility, she must also be connected to at least one other red agent, which completely characterizes the worst-case topology and equilibrium assignment. In particular, in this topology there exists a node of degree which is connected to leaf nodes, while the assignment is such that is occupied by and all leaf nodes are occupied by blue agents; see Figure 2. Hence, we have

On the other hand, the optimal assignment is obtained after swapping with one of the blue agents. This yields

and the price of anarchy is

which is maximized to for .

Next, we assume that all agents have positive utility. Since is an equilibrium, for every red-blue pair of agents, at least one of them has no incentive to swap positions. Let be a pair of red-blue agents, and assume that does not want to swap. If and are not neighbors according to , then it must be that

Otherwise, if and are neighbors according to , it might be the case that , in which case and . Assume that the blue agent has red neighbors besides , and blue neighbors. Then, , and

(1)

where the inequality follows since is decreasing in and . Therefore, in any case we have that , for every red-blue pair of agents and . Since there are distinct red-blue pairs of agents, and each agent participates in exactly such pairs, by summing over all these inequalities, we obtain that

The bound on the price of anarchy now follows by observing that the maximum possible social welfare is , since the utility of each agent is at most . ∎

We remark that even though the upper bound of for the case where each types consists of the same number of agents (and every agent has positive utility at equilibrium) is probably not tight, one cannot expect to improve it using the same technique. In particular, to prove this upper bound, we focused on an arbitrary pair of agents , and used the equilibrium definition, according to which at least one of these agents does not want to swap positions. Then, we were able to show that . We now argue that this inequality is actually tight, and hence to improve the upper bound one needs to argue in more detail about the structure of the equilibrium. Consider a variant of the topology depicted in Figure 2 in which every leaf node is connected to another leaf node. Then, for the depicted assignment , the red agent has utility , while any blue agent has utility exactly . Hence, the sum of the utility of the red agent and the utility of any blue agent is almost as becomes large.

We continue by showing that, surprisingly, for three types or more, the social price of anarchy can be unbounded, even for the special case of equal number of agents per type.

Figure 3: The topology and the equilibrium assignment of the -swap game considered in the proof of Theorem 4.2 for . Here, , and .
Theorem 4.2.

For every , the social price of anarchy of -swap games can be unbounded, even when there is an equal number of strategic agents per type.

Proof.

Consider a -swap game with agents such that there are exactly two agents of each of the types . The topology consists of nodes that form a cycle, and each node , is also connected to an auxiliary node ; see Figure 3 for the topology and the equilibrium assignment discussed in the following for .

Let be the assignment according to which node is occupied by an agent of type , while node is occupied by an agent of type , where the subscripts are mod . This assignment is clearly an equilibrium since there exists no pair of agents that would like to swap positions, and every agent has zero utility. In particular, observe that agent of type would like to move only to node in order to connect with the agent who is also of type . However, the agent of type has no incentive to move to node since the other agent of type is at node . Now, consider agent of type who is connected only to an agent of type located at . The only agent that would like to swap positions with is who is of type . However, such a swap cannot increase the utility of since agent is of type . Therefore, is an equilibrium with .

On the other hand, consider the assignment according to which nodes and are occupied by the two agents of type , for every . Since every agent has now positive utility, , and the social price of anarchy is unbounded. ∎

Next, we turn our attention to the social price of stability and show a constant lower bound for -swap games.

Figure 4: The topology and the best equilibrium assignment for the -swap games considered in the proof of Theorem 4.3.
Theorem 4.3.

The social price of stability of -swap games is at least .

Proof.

Let be a parameter, and consider a -swap game with red and blue agents. The topology is a tree with a root node , which is connected to two nodes and , as well as to a set of leaf nodes. Moreover, node is connected to a set of leaf nodes, and node is connected to a set of more leaf nodes. The topology and the best equilibrium assignment (which we discuss below) are depicted in Figure 4.

We will now argue about the structure of any equilibrium for this particular swap game. Without loss of generality, we assume that the root node is occupied by a red agent, and switch between a few cases depending on the number of blue agents that occupy the lead nodes of set that are directly connected to .

First, assume that there are at least blue agents at the nodes of set . Then, there can be at most red agents at the nodes of , which means that the remaining at least red agents need to occupy nodes of the - and -subtrees. Since any of these subtrees have a total of nodes, at least one of these red agents, say agent , must be connected to at least one blue agent. Clearly, such an assignment cannot be an equilibrium since agent and any of the blue agents at the nodes of have incentive to swap positions to increase their utilities from strictly smaller than and to and positive, respectively.

Second, assume that there are exactly blue agents at the nodes of set , and hence the remaining nodes of are occupied by red agents. Then, it is easy to verify that the only equilibrium assignment (up to symmetries) is such that all nodes of the -subtree are occupied by red agents, and all nodes of the -subtree are occupied by blue agents. The social welfare of this equilibrium is

Third, assume that the number of blue agents at the nodes of set is between and . Then, there are at least red agents at the nodes of . Since any of the - and -subtrees have a total of nodes, at least one of the remaining at most red agents, say agent , must necessarily be connected to some blue agent. As in the first case, such an assignment cannot be an equilibrium since agent and any of the blue agents at the nodes of have incentive to swap positions to increase their utilities from strictly smaller than and to and positive, respectively.

Finally, assume that all nodes of are occupied by red agents, and there is only one remaining red agent , who will inevitably be connected to some blue agents. No assignment according to which occupies a leaf node of (or ) can be an equilibrium, since and the blue agent (or ) have incentive to swap positions and increase their utilities from and strictly smaller than to positive and , respectively. Hence, in any equilibrium assignment , agent occupies either node or node . The social welfare is

Now consider the assignment according to which the red agents occupy node , all nodes of , and one node of , while all other nodes are occupied by blue agents. The social welfare of this assignment is

Therefore, the social price of anarchy is at least , which tends to as tends to infinity. ∎

For -swap games with topology that is a -regular graph (in which all nodes have degree equal to ), we show an upper bound of on the social price of stability by exploiting a potential function, similar to the one defined by Chauhan et al. (2018) and Echzell et al. (2019) to show the existence of equilibria in such games.

Theorem 4.4.

The social price of stability in -swap games with topology that is a -regular graph is .

Proof.

Echzell et al. (2019) showed that for -swap games with a -regular topology,

is a potential minimization function. Using similar arguments, we can show that the complement,

is a potential maximization function. Consider any pair of agents such that is of type and is of type , with . Since and swap positions if and only if they can both increase their utility, and since for any assignment , we have that

Any agent has one less friend in than in , and hence

On the other hand, any agent has one more friend in than in , and hence

For any other agent, the friends they have as neighbors have not changed. Therefore, we can now easily see that

as desired.

Now, observe that by the definition of the utility of each strategic agent and the fact that the topology is -regular, for any assignment , we have that

(2)

Let be an optimal assignment. If is an equilibrium, then the social price of stability is . Otherwise, we let the strategic agents play and swap positions until they reach an equilibrium . Since is a potential maximization function, we have that , and by (2), we obtain

and the bound follows by rearranging terms. ∎

Next, we focus on the problem of computing assignments of high social welfare. Observe that whether the agents are allowed to pairwise swap positions or jump to empty nodes has no effect in the complexity of this problem, and hence we already known that it is NP-complete by the work of Elkind et al. (2019). However, one of their main assumptions is that the topology is a graph with strictly more nodes than agents (so that there are empty nodes where the agents can jump to). Consequently, their proof does not cover our case, where the topology consists of a number of nodes that is exactly equal to the number of agents. The proof of our next theorem is fundamentally different and subsumes that of Elkind et al. (2019) for ; for , we were unable to prove the hardness of the problem.

Theorem 4.5.

For every , given a rational number , it is NP-complete to decide whether there exists an allocation that has social welfare at least .

Proof.

Membership in NP is immediate: given an assignment, we can sum up the utilities of the strategic agents and check whether the social welfare is at least . To prove NP-hardness, we give a reduction from an NP-complete variant of the min-cut problem with additional cardinality constraints on the size of the subsets, to which we refer as the Equal-Min-Cut problem Garey et al. (1974). An instance of Equal-Min-Cut consists of a graph , two distinguished nodes , and an integer . It is a yes-instance if and only if there exist disjoint subsets of nodes and such that , , , and . Without loss of generality, we assume that is an even number, and by convention we denote an edge as to simplify our notation.

Given an instance of Equal-Min-Cut with , we construct an instance of our social welfare maximization problem as follows:

  • There are strategic red and strategic blue agents.

  • The topology consists of with additional nodes and edges. Let and be two auxiliary nodes, and define and . Let for every , and . Also, let be a set of nodes for every . Then, is such that and . Observe that in , every node has degree exactly equal to .

  • The nodes and are occupied by stubborn red agents, and are occupied by stubborn blue agents, and all nodes in are occupied by stubborn green agents.

  • Finally, let .

For any assignment and node , let denote the type of the agent occupying node . We will show that the social welfare of is decreasing in the number of edges of that are occupied by agents of different types. We have

Since , , , and , it follows that for every ,

Furthermore,

Therefore, we obtain

(3)

Now, assume that the input instance of Equal-Min-Cut is a yes-instance, and let be the satisfying partition. Let be such that the strategic red agents occupy the nodes of and the strategic blue agents occupy the nodes of . Then, by definition we have that , and by (3) we obtain

Conversely, assume that there exists an assignment with . Let consist of the nodes occupied by strategic red agents, and let consist of the nodes occupied by strategic blue agents. Then, , and since there is an equal number of strategic red and blue agents, we also have that . By (3), it is

and consequently

as desired. ∎

5 Degree of Integration

We now investigate whether equilibrium assignments can be diverse, by bounding the price of anarchy and stability in terms of the degree of integration; recall that this benchmark counts the number of agents who are exposed, i.e., have at least one neighbor of a different type. As in the previous section, we again focus on games with strategic agents only.

We start by showing that the integration price of anarchy of -swap games is